# A problem about a non-linear differential equation

Given the equation $$x''(t)+x'(t)+x^3(t)=0$$ with $$x(0)=1$$ and $$x'(0)=0$$ how to prove that

1. $$\mathop {\lim }\limits_{t \to + \infty } x\left( t \right) = 0$$.
2. For every $$t>0$$ it is $$x(t)>0$$.

the second question is what I really need. This problem is from G. Gallavotti "The elements of Mechanics"

Consider the energy function $$E=\frac12 x'^2+\frac14x^4$$, then the time derivative of that is $$E'=x'(x''+x^3)=-x'^2\le 0.$$ The only solution that has $$x'=0$$ constantly is the stationary solution $$x=0$$. All other solutions continuously lose energy and fall towards the state $$E=0$$.
The second question is more complicated, you have to show that the friction is over-critical, so that no oscillation occurs. The equation $$x''+x^3=0$$ without friction is conservative and oscillates along the level curves of $$E$$, for a small friction coefficient this oscillation persists, giving a spiral in phase space.
As this plot of $$x(t)$$ shows, a friction coefficient of $$0.8$$ still crosses the zero line, any proof method for 2. must be quite specific for $$c>0.9$$.
• In zooms for $c\in[0.9,1]$ it looks like it has a boundary layer. You can write it as system $x'+cx=v$, $v'=-x^3$, where in some sense the first is the "fast" and the second the "slow" equation. Then when $x$ is small enough, $v$ is slow moving while the first equation restores its equilibrium at $x=v/c$. Insert this approximation into the second to get $cx'=-x^3\implies x^{-2}(t)-x^{-2}(t_1)=2(t-t_1)/c$ as long-time behavior. I'm just not sure how to prove how exact this is, what "small enough" is, and how to connect to the initial piece. – Lutz Lehmann Feb 17 at 8:47